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Triangle T(n,k) read by rows: Padovan factorial ratios T(n,k) = round(c(n)/(c(k)*c(n-k))) where c are partial products of a generalized Padovan sequence with multiplier q = 2defined in comments.

Start from the generalized Padovan sequence A159284 and its partial products c(n) = 1, 1, 1, 1, 3, 9, 45, 405, 4455, 84645, 2454705, ... . Then T(n,k) = round( c(n)/(c(k)*c(n-k)) ).

Row sums are 1, 2, 3, 4, 11, 26, 87, 380, 1707, 10490, 79955, ...

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Triangle tT(n,k) read by rows: Padovan factorial ratios round(c(n)/(c(k)*c(n-k)) ) where c are partial products of a generalized Padovan sequence with multiplier mq = 2.

Start from the generalized Padovan sequence A159284 and its partial products c(n) = 1, 1, 1, 1, 3, 9, 45, 405, 4455, 84645, 2454705, ... . Then tT(n,k) = round( c(n)/(c(k)*c(n-k)) ).

Row sums are 1, 2, 3, 4, 11, 26, 87, 380, 1707, 10490, 79955, ...

Note that rows n>=14 contain fractions. R. J. Mathar, Jul 05 2012

G. C. Greubel, <a href="/A172358/b172358.txt">Rows n = 0..50 of the triangle, flattened</a>

T(n, k, q) = round(c(n,q)/(c(k,q)*c(n-k,q)), where c(n,q) = Product_{j=1..n} f(j,q), f(n, q) = f(n-2, q) + q*f(n-3, q), f(0,q)=0, f(1,q) = f(2,q) = 1, and q = 2. - G. C. Greubel, May 09 2021

Triangle begins as:

1;

1, 1;

1, 1, 1;

1, 1, 1, 1;

1, 3, 3, 3, 1;

1, 3, 9, 9, 3, 1;

1, 5, 15, 45, 15, 5, 1;

1, 9, 45, 135, 135, 45, 9, 1;

1, 11, 99, 495, 495, 495, 99, 11, 1;

1, 19, 209, 1881, 3135, 3135, 1881, 209, 19, 1;

1, 29, 551, 6061, 18183, 30305, 18183, 6061, 551, 29, 1;

Clearf[n_, q_]:= f[n, q]= If[n<3, Fibonacci[n], f[n-2, q] + q*f, c, a, t[n-3, q]];

f[0, a_] := 0; f[1, a_] := 1; f[2, a_] := 1;

c[n_, q_]:= Product[f[j, q], {j, n}];

fT[n_, a_k_, q_] := fRound[c[n, aq] = f/(c[n - 2, ak, q] + a*fc[n - 3, ak, q])];

c[n_, a_] := If[n == 0, 1, Product[f[i, a], {i, 1, n}]];

t[n_, m_, a_] := c[n, a]/(c[m, a]*c[n - m, a]);

Table[Table[Table[t[n, m, a], {m, 0, n}], {n, 0, 10}], {a, 1, 10}];

Table[Flatten[Table[Table[t[n, m, a], {m, 0, n}], {n, 0, 10}]], {a, 1, 10}]

Table[T[n, k, 2], {n, 0, 12}, {k, 0, n}]//Flatten (* modified by G. C. Greubel, May 09 2021 *)

(Sage)

@CachedFunction

def f(n, q): return fibonacci(n) if (n<3) else f(n-2, q) + q*f(n-3, q)

def c(n, q): return product( f(j, q) for j in (1..n) )

def T(n, k, q): return round(c(n, q)/(c(k, q)*c(n-k, q)))

flatten([[T(n, k, 2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 09 2021

Cf. A172353 (q=1), this sequence (q=2), A172359 (q=4), A172360 (q=5).

Definition corrected to give integral terms by G. C. Greubel, May 09 2021

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Fifth (third cubic) type of beta integer triangle sequence: a=2;f(n,a)=f(n-2,a)+a*f(n-3,a);c(n,a)=If[n == 0, 1, Product[f(i, a), {i, 1, n}]];t(n,m,a)=c(n, a)/(c(m, a)*c(n - m, a))

Triangle t(n,k) read by rows: Padovan factorial ratios c(n)/(c(k)*c(n-k)) where c are partial products of a generalized Padovan sequence with multiplier m=2.

Start from the generalized Padovan sequence A159284 and its partial products c(n) = 1, 1, 1, 1, 3, 9, 45, 405, 4455, 84645, 2454705... Then t(n,k) = c(n)/(c(k)*c(n-k)).

Row sums are: 1, 2, 3, 4, 11, 26, 87, 380, 1707, 10490, 79955,...

{1, 2, 3, 4, 11, 26, 87, 380, 1707, 10490, 79955,...}.

Note that rows n>=14 contain fractions. R. J. Mathar, Jul 05 2012

a=2;

f(n,a)=a*f(n-1,a)+a*f(n-3,a);

c(n,a)=If[n == 0, 1, Product[f(i, a), {i, 1, n}]];

t(n,m,a)=c(n, a)/(c(m, a)*c(n - m, a))

{1},

1;

{1, 1},;

{1, 1, 1},;

{1, 1, 1, 1},;

{1, 3, 3, 3, 1},;

{1, 3, 9, 9, 3, 1},;

{1, 5, 15, 45, 15, 5, 1},;

{1, 9, 45, 135, 135, 45, 9, 1},;

{1, 11, 99, 495, 495, 495, 99, 11, 1},;

{1, 19, 209, 1881, 3135, 3135, 1881, 209, 19, 1},;

{1, 29, 551, 6061, 18183, 30305, 18183, 6061, 551, 29, 1};

cf. A010048

nonn,tabl,unedless

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_Roger L. Bagula (rlbagulatftn(AT)yahoo.com), _, Feb 01 2010